******************************Geometry Problem of the Week, November 11-15

You might have to read this a couple of times and draw a picture before itmakes any sense!

A chord of a circle is the hypotenuse of an isosceles right triangle whoselegs are radii of the circle. The radius of the circle is 6 times thesquare root of 2. What is the length of the minor arc subtended by thechord?

What if it were an equilateral triangle instead of just isosceles?

What if I told you that the non-base angle (so the angle at the vertex atthe center of the circle) was x degrees? Then what would the answer be?

This problem has lots of parts, but I think if you really understand thefirst part, the second two won't be too bad. Make sure you really explainyour answer!

******************************

A pretty good job this week! 80 correct solutions were submitted, and 35incorrect. The majority of the incorrect solutions were missing the thirdpart - it is sometimes hard to generalize something even though you usedit correctly to solve two other parts.

One common mistake was to get the third part backwards. That is, dividingby x/360 instead of multiplying by it. This is after doing exactly thefirst thing in the first two parts. You need to look very carefully atwhat you did in the specific cases.

Several people used a formula for arc length that requires the length ofthe chord and the angle. It works okay for the first two parts, but makesthethird part harder, I think.

For accuracy, leave everything in terms of pi and sqrt2 until the veryend. You might also find that a lot of stuff cancels out before you'redone, andto me, the less numbers, the better! In fact, with a problem like this, Ilike to see the answers in terms of pi and sqrt2 _and_ as just numbers. Itmakes the answer easier to follow and connects it with the circle and theoriginal radius.

This problem is a good time to work on writing good equations. Explaincarefully what you are going to do, and where the numbers come from, butwhen it comes to presenting the numbers, write some equations.

I've highlighted two solutions this week. Thomas Bereknyei from HillviewJunior High gave a simple explanation. Greg Moore and Shawn Phelenfrom Shaler Area High School pretty much explained it perfectly. There arealso a number of good solutions in the full list of solutions, so be sureyou read them over.

The following students submitted correct solutions this week. A few of thesolutions are highlighted below. Most of the rest of the solutions arealso available.

If the legs are the radii of the circle, then the hypotenuse must be the chord, and the central angle must be 90 degrees.The minor arc between two radii is equal to the same fraction of the circumfrence as the fraction created by the angle over 360 degrees.Therefore, if the radius is equal to 6 times the square root of 2, the circumfrence is equal to 12pi times the square root of 2 (2*pi*R).The central angle (90 degrees) over 360 degrees equals one fourth, which, multiplied by the circumfrence, equals 3pi times the square root of 2.If the triangle was equilateral, then all three angles, including the central angle, would be 60 degrees. 60/360 times the circumfrence equals 2pi square root of 2.Finally, if the angle was x degrees, then the arc would equal (x / 360)(12pi times the square root of 2).However, if the angle was greater than 180 degrees, the minor arc would be (360 - x)/360 * 12pi*square root of 2.

We shall call the point at the center of the circle point A, thecircumference of the circle shall be C and the radius shall be R.R = 6 times the square root of 2. C=2 times pi times R. So,C=12 times the square root of 2 times pi. The total number ofdegrees in a circle is 360, and since the two legs of thetriangle are the radii of the circle and angle A is 90 degrees,the minor arc subtended by the hypotenuse of the triangle (alsothe chord) is one quarter of C, which is 3 times root 2 times pi.If the triangle were equilateral, then it would also beequiangular, so angle A = 60 degrees. Since 60 is 1/6 of 360,the minor arc subtended by the chord would be 1/6 of C, which is2 times root 2 times pi.If angle A = x, then the minor arc subtended by the chord wouldbe equal to (12 times root 2 times pi) divided by 360/x.

Josh GrochowGrade 8Paul NassGeorgetown Day SchoolNovember 11-15, 1996

***********************************************

From: NRGM22A@prodigy.com

From: Katherine WaltherGrade: 9School: John Glenn Junior High

To figure this problem out, all you need to know is the measure of thevertex angle of the triangle and the radius ofthe circle. The measure of the minor arc created by the chord is equal to the measureof the vertex angle over 360 times the circumference of the circle.

Since the triangle in the first part of the problem is a righttriangle, the angle of the arc subtended by the chord is 90degrees or pi/2 radians. Since the length of the arc is 6 timesthe square root of 2, the length of the arc is 3 times thesquare root of two times pi. If the triangle were isosceles,the angle subtended would be 60 degrees or pi/3 radians.Therefore, the length of the arc would be 2 times the squareroot of two times pi. Were the angle x degrees, the subtendedarc would have a length of the square root of two times x timespi over 30. These calculations all are based on the fact thatthe length of an arc is equal to the radius of the circle timesthe represented by the arc.

The circumference of the circle with radius of 6SQRT(2) is 2(PI)(6SQRT(2))=12SQRT(2)PI.

(PART 1) Since 90 degree is 1/4 of 360 degree, the lentgth of the minor arc is (1/4)(12SQRT(2)PI) = 3SQRT(2)PI.

(PART 2) If the triangle is equilateral, then the vertex angle is 60 degree which is 1/6 of 360 degree. Therefore, the length of the minor arc is (1/6)(12SQRT(2)PI) = 2SQRT(2)PI.

(PART 3) If the vertex angle is x degree which is (x/360) of 360 degree. Therefore, the length of the minor arc is (x/360)(12SQRT(2)PI) = xSQRT(2)PI/30.

***********************************************

From: arthur@iolani.honolulu.hi.us

From: Jason YeungGrade: 10School: Iolani School, Honolulu, Hawaii

Answer: 3 (sqrt 2) * pi 2 (sqrt 2) * pi x (sqrt 2) * pi / 30

The isosceles right triangle's right legs are the radii of the circle, thus the angle is a central angle, which implies if theangle is 90 degrees, then the angle of the arc is also 90 degrees.The circum. of the circle is 12 sqrt 2 pi. The arc is 90 degrees over 360 degrees (the whole circle), or 1/4 the circum. of the circle. Therefore, the length of the minor arc is 3 (sqrt 2) * pi.

If the triangle is equilateral instead, then the central angle would be 60 degrees, and the arc would be 1/6 of the total circle.Therefore, the length of that minor arc is 2 (sqrt 2) * pi.

If the non base angle is x degrees, then the central angle would be x degrees, and the arc would be x/360 of the total circle.Therefore, the length of that arc is x (sqrt 2) * pi / 30.

1. First I found that 8.5 was the radii. Then using the central angle, Ifound the ratio of it to the circumfenrce was 90/360=1/4. Next I found the circumference by using pi*D, 53.4, then multiplied it by the 1/4 ratio to get the arc. The arc was 13.35.

2. I followed all the same steps in the second one that I used in the first. However the central angle was 60 so the ratio was 60/360=1/6. The arc =8.9.

3. Because the central angle is x, it could be any size. The ratio would be x/360 and the arc would be x/360*53.4.

A circle has 360. The radius of this one rounds off to 8.5. An isoscelesright triangle has 2 = sides. One < is 90. 360 divided by 90 is 4 so the isosceles triangle is 1/4 of the circle. The we got the circumference of the whoe circle = 53.38. Since the triangle takes up 1/4 of the circle, we difided 53.38 by 4 and got 13.345, so the minor arc = 13.345.

An equialteral triangle has 180 and each < is 60. We divided 360 by 60 to see how many triangles would fit in the circle. We got 6. We got c=53.4 and divided by 6. This arc = 8.9.

For the last part we would divide 360 by x since for the 1s one we divided 360 by 90 and the 2nd one we divided by 60 then we divide the 53.38 by 360/x.

A circle is 360. The radius here is 8.5. Divide the 360 by 90=4. So the triangle is 1/4 of the circle. C=53.38. Since the triangle takes up 1/4of the circle divide 53.38 by 4 = 13.3, so the minor arc is 13.3.

We divided 360 by 60 for the equilateral triangle and we got 6. We got the circumference to be 53.4. Then we divided by 6 and got 8.9 to be the answer.

Fopr the last part, we would divide 360 by x and divide c by the answer.

If the angle of an isosceles triangle fitted into circle = 90 and circle = 360 then the angle is a quarter of the circle. In relation, the subtended arc is a quarter of the circumf=1/4(53.3 = 13.326.

If the angle of an equilateral triangle, fitted into circle=60 and the circle equals 360 then the < is a sixth of the circle. In relation, the subtended are is a sixth of the circumference = 1/6(53.3)=8.883

The non-base < in which the hyp. is a cord of the circle, measures 90. Since there are 360 in a circle, this angle is 1/4 of the circle. Therefore, the subtended arc is 1/4 of the circumference. C=2*pi*r. 4=8.48. Therefore c=53.28745703. 1/4 of the circumf. = 13.32189176

In the circle with 60 central angle. This is 1/6 of the circle. Therefore, the measure of the subtended arc is 1/6 of the circumference. The circumf. is the same, so 1/6(53.28756703)=8.881261172.

If the measure of the angle is x, then the subtended arc is x/360*circumference of the circle.

The length of the minor arc subtended by the chord of an isoscelesin a circle with a leg of 6 times the square root of 2 is 3 pi timesthe square root of 2, because the circumfrence of the circle is 12 pi times the square root of 2, or 2 times the radius (6 times the square root of 2) times pi. Since the non-base angle of the isosceles triangle is 90, then the length of the arc subtended bythe hypotenuse is one fourth the entire circumfrence of the circle,because 90 degrees is one fourth of 360, the total degrees in a circle.The length of the minor arc subtended by the hypotenuse of an equilateralin a circle with a side of 6 times the square root of 2 is 2 pi timesthe square root of 2. Since the non-base angle of the equilateral triangle is 60 (each angle is 60 degrees in an equilateral triangle)then the length of the arc subtended by the hypotenuse of the equilateral triangle is one sixth of the entire circumfrence of thecircle because 60 degrees is one sixth of 360.The length of the minor arc subtended by the chord of the circle is X times pi times the square root of 2, the quantity over 30, whereX is the measure of the angle at the vertexat the center of the circle, because the circumfrence needs onlyto be divided by the times the angle goes into 360, so the circumfrenceof the circle divided by the quantity of 360 divided by the angle at the center of the circle.

The minor arc subtended by the hypotenuse of the right isoscelestriangle has a length of about 13.33. The minor arc subtended by the chord in the equilateral triangle has a length of about 8.886,while the central angle formed by the legs of the right triangle is equal to 90 degrees (x=90 degrees).

I obtained this information by first diagraming the problems.Since the right anle of the triangle is the central angle of the circle, I divided 90 degrees by 360 degrees (There are 360 degrees in a circle). This gave me a fraction of 1/4. Thus, the central angle occupies 1/4 of the circle. Corespondingly, the minor arc subtendedby this trianle has a length equal to 1/4 of the circumference ofthe circle. Thus: 1/4*C=M, where M= the minor arc subtended by the chord C= the circumference of thecircle.

Since the circumference of any circle is equal to the radiusmultiplied by 2 and pi (about 3.1416), the total length of the minorarc can be expressed in the folowing formula:

1/4*(3.1416*2*6*(the square root of 2))=M

When this is reduced one gets: 3.1416*3*(the square root of 2)=M

The answer becomes about 13.33.

I applied the same method to the equilateral triangle.Since the sum of any trianle's angles must be equal to 180, the equilateral triangle must be composed of three 60 degree angles(180/3= 60). 60 degrees divided by 360 degrees is equal to 1/6, so the central angle must comprise 1/6 of the whole circle. Corespondingly, the minor arc must also have a length equal to 1/6of the circle's circumference.

This formula will give you the length of the arch formed by theradii of any central angle. I determined this by noting that any circle has360 degrees, and any central angle in the circle forms some portion of thecircle. Since this portion of the circle can be expressed by x/360, thenone can multiply x/360 by the circumference to determine the lenght of the arc.

***********************************************

From: Lishack@sasd.k12.pa.us

From: Bob Young and Christian PaulGrade: 10School: Shaler Area High School

1. If the radius of the circle is 6 square-roots of 2 then, the circumference of the circle is 53.31. In the first part this should be divided by 4, since four triangles can fit into this circle. This would be 13.33.2. Since the circumference is the same, 53.31, then all we had to do wasdivide this by 6 instead of four, since 6 triangles fit into the circle. Theanswer is 8.89.3. If you don't know what degree the triangle is, use this formula for your answer: 2*Pi*R(x/360).

To find our answers we used this formula:Degrees/180*Pi*rFor the isoceles triangle the degrees was 90, so we put it intothe formula and got 13.3.For the equilateral triangle the degrees was 60, so we put itinto the formula and got 8.89.Then it asked to find degrees which was x, so we got 26.7x/180.

First we defined all of the unknown terms. Then we came to concluusions on how to draw the drawing. We inscribed an isoscelesright triangle into a circle. Since we know the legs of the triangle are the radii of the circle, they are both 6*the square root of two, which is 8.49. To find the length of the minor arc, we used the formula (q/180)*pi*r, where q is equalto the degree of the triangle. Since the first triangle is a right triangle it is 90 degrees. The final answer is 13.34 rounded to the nearest hundreth. A equilateral triangle is 60 degrees, we put it into the same formula to find the minorarc which was 8.89. When the angle is "x", using the same formula(q/180)*pi*r), the answer in terms of x was (26.67x/180).